Find the area bounded by y(x+2)=x^4 , x=0,y=0 and x=3


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Question Number 61986 by necx1 last updated on 13/Jun/19

$F i n d t h e a r e a b o u n d e d b y y (x + 2) = x^{4},$ $x = 0, y = 0 a n d x = 3$
	Answered by mr W last updated on 13/Jun/19

	$y = \frac{x^{4}}{x + 2}$ $A = \int_{0}^{3} \frac{x^{4}}{x + 2} d x$ $A = \int_{0}^{3} \frac{(x + 2) (x - 2) (x^{2} + 4) + 16}{x + 2} d x$ $A = \int_{0}^{3} [(x - 2) (x^{2} + 4) + \frac{16}{x + 2}] d x$ $A = \int_{0}^{3} [x^{3} - 2 x^{2} + 4 x - 8 + \frac{16}{x + 2}] d x$ $A = {[\frac{1}{4} x^{4} - \frac{2}{3} x^{3} + 2 x^{2} - 8 x + 16 \ln (x + 2)]}_{0}^{3}$ $A = [\frac{1}{4} 3^{4} - \frac{2}{3} 3^{3} + 2 \times 3^{2} - 8 \times 3 + 16 \ln \frac{3 + 2}{2}]$ $A = [- \frac{15}{4} + 16 \ln \frac{5}{2}] = 10.91$
	Commented by necx1 last updated on 13/Jun/19

	$T h a n k y o u s o m u c h .$
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